The functional differential equation $f'(x) = f(f(x))$

Question

In Maths 505 I found an interesting functional differential equation (FDE) problem (https://youtu.be/C6fZVwqhbnE?si=Trxk-KYNmQUeEPCv) which asks for the solution of the equation

$$f'(x) = f(f(x))\tag{1}$$

No further conditions were given.

I had never before studied this type of equations. Here is what I did so far.

The first attempt was suggested by the author of the problem: make the power ansatz with two parameters

$$f(x) = a x^{b }\tag{2}$$

Then we have $f'(x) = a b x^{b-1}$ and $f(f(x)) = a( a x^{b })^{b} = a^{1+b} x^{b^2}$ from which we deduce $b^2=b-1$ and $b=a^{b}$.

These equations have the solutions

$$b_{\pm} = \frac{1\pm i \sqrt{3}}{2}=e^{\pm \frac{i \pi}{3}}\tag{3a}$$

and

$$a=b^{1/b}\tag{3b}$$

So that we have found two solutions to $(1)$

$$f_\pm(x) = b_\pm^{1/b_\pm} x^{b_\pm}\tag{3c}$$

Expanding this using Euler's formula we can calculate the real and imaginary parts of $f(x)$. This shows that we have obtained two complex functions of the argument $x$ which we assumed he to to be real.

We can study the properties of the solution, plot graphs etc. but one problem appears - and remains - urgently: a dfferential question of first order must have one free parameter to adapt the solution to an initial condition.

Here I am stuck since I can't accomodate an arbitrary constant in the solution found by a power ansatz.

Second approach:

Let us assume the initial condition

$$f(1)=1\tag{4}$$

This simplifies the arithmetic considerably. Indeed, we find $f'(1) = f(f(1)) = f(1) = 1$, and next $f''(x) = \frac{d}{dx} f(f(x)) = f'(f(x))\cdot f'(x) = f(f(f(x))) \cdot f(f(x))$ giving $f''(1) = 1$, $f'''(1) = 2$ etc.

Continuing this procedure (conveniently done in Mathematica) we find for $f(1)$ and the first nine derivatives of $f$ at $x=1$ the strongly increasing series

$$f^{(k)}(1)|_{k=0}^{k=9} = \{1,1,2,7,37,269,2535,29738,421790,7076459\}\tag{5}$$

This series is contained in OEIS as https://oeis.org/A001028 and reads A001028 E.g.f. satisfies A'(x) = 1 + A(A(x)), A(0)=0.

This remark confirms our suspection: "The e.g.f. is diverging". Hence we have not obtained a valid series expansion around $x=1$.

Here I'm stuck again.

Hopefully someone can solve the problem with, a proper initial condition, say f(1) = 1/2$.

EDIT

I am grateful to a user (who deleted the contribution shortly after publication) who notified me that the problem was studied earlier by Alex Jones in: https://www.quora.com/How-can-I-solve-f-x-f-f-x where my first attempt was described. The general problem was not attacked there.

Answer 1

There are two closed form solutions:

$$\displaystyle f_1(x) = e^{\frac{\pi}{3} (-1)^{1/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$ $$\displaystyle f_2(x) = e^{\frac{\pi}{3} (-1)^{11/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$

The solution technique can be found in this paper.

For a general case, solution of the equation

$$f'(z)=f^{[m]}(z)$$

has the form

$$f(z)=\beta z^\gamma$$

where $\beta$ and $\gamma$ should be obtained from the system

$$\gamma^m=\gamma-1$$ $$\beta^{\gamma^{m-1}+...+\gamma}=\gamma$$

In your case $m=2$.